What is the derivative of (x+4/(9x^2e^x)) ?

The derivative of (x+4/(9x^2e^x)) is -(x^2+5x+8)/(9e^xx^3)

What is the first derivative of (x+4/(9x^2e^x)) ?

The first derivative of (x+4/(9x^2e^x)) is -(x^2+5x+8)/(9e^xx^3)

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\frac{d}{d x} (\frac{x + 4}{9 x ^{2} e ^{x}})

- \frac{x ^{2} + 5 x + 8}{9 e ^{x} x ^{3}}

Solution steps

\frac{d}{d x} (\frac{x + 4}{9 x ^{2} e ^{x}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{9} \frac{d}{d x} (\frac{x + 4}{x ^{2} e ^{x}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{1}{9} \cdot \frac{\frac{d}{d x} ( x + 4 ) x ^{2} e ^{x} - \frac{d}{d x} ( x ^{2} e ^{x} ) ( x + 4 )}{( x ^{2} e ^{x} ) ^{2}}

\frac{d}{d x} (x + 4) = 1

\frac{d}{d x} (x^{2} e^{x}) = 2 x e^{x} + e^{x} x^{2}

= \frac{1}{9} \cdot \frac{1 \cdot x ^{2} e ^{x} - ( 2 x e ^{x} + e ^{x} x ^{2} ) ( x + 4 )}{( x ^{2} e ^{x} ) ^{2}}

Simplify

\frac{1}{9} \cdot \frac{1 \cdot x ^{2} e ^{x} - ( 2 x e ^{x} + e ^{x} x ^{2} ) ( x + 4 )}{( x ^{2} e ^{x} ) ^{2}} : - \frac{x ^{2} + 5 x + 8}{9 e ^{x} x ^{3}}

= - \frac{x ^{2} + 5 x + 8}{9 e ^{x} x ^{3}}

What is the derivative of (x+4/(9x^2e^x)) ?
The derivative of (x+4/(9x^2e^x)) is -(x^2+5x+8)/(9e^xx^3)
What is the first derivative of (x+4/(9x^2e^x)) ?
The first derivative of (x+4/(9x^2e^x)) is -(x^2+5x+8)/(9e^xx^3)